Let $k$ be a field.
$\exists \xi \subset k^*$, $o(\xi) = n$.
$a \in k$, $\forall d>1$, $d \mid n$, $\not\exists b \in k$ such that $a = b^d$.
Need to find the Galois group of $x^n - a$.
I've seen this solution, but it's not helped me. I appreciate any ideas for solving this problem.
For a field $k$ of characteristic not dividing $n$ and containing the group $\mu_n$ of $n$-th roots of unity, Kummer's theory describes explicitly the finite abelian extensions $K/k$ of exponent $n$ (i.e. finite Galois extensions whose groups are abelian, killed by the exponent $n$), more precisely $\mathrm{Gal}(K/k) \cong \mathrm{Hom} (A,\mu_n)$, where the "Kummer radical" $A$ is the subgroup of $k^*/(k^*)^{n}$ s.t. $K$ is obtained by adding to $k$ the $n$-th roots of the elements of $A$. See e.g. S. Lang's "Algebra", chap. VIII, §6.
In your special case, $A$ is just the subgroup of $k^*/(k^*)^{n}$ generated by the class $\left<a\right>$ of an element $a\in k^*$, hence the relevant Kummer extension is cyclic of exponent $n$, with group $G\cong \mathrm{Hom}(\left<a\right>, \mu_n)$. Actually, by duality, the order of $G$ is just equal to the order $d$ of the group $\left<a\right>$, or equivalently the order of the class of $a$ mod $(k^*)^{n}$. By definition, this $d$ is characterized by the conditions given in your post (for details, see op. cit., thm. 11).
NB: Outside the hypotheses of Kummer's theory, there are criteria for the irreducibility of $X^n - a$, see op. cit., thm. 16. The determination of the Galois group of this polynomial must then follow the usual path: the splitting field is $L=k(\mu_n, \sqrt [n]a)$, and one knows $L/k(\mu_n)$ by Kummer's theory if $k$ has characteristic $p$ not dividing $n$; the technical point remains to describe the Galois group over $k$. If $p=n$, the Artin-Schreier theory (op. cit., thm. 15) describes $L/k$, replacing $p$-th powers by the operator $x \to x^p-x$. For $n=p^a$, there is a more complicated analog of Artin-Schreier using the so called "Witt vectors" (op. cit., chap. VIII, ex. 25) .
